# Time until distance gets equal to X between two objects moving in opposite direction

Consider two people moving in opposite direction with speeds **U meters/second** and **V meters/second** respectively. The task is to find how long will take to make the distance between them X meters.

**Examples:**

Input:U = 3, V = 3, X = 3

Output:0.5

After 0.5 seconds, policeman A will be at distance 1.5 meters

and policeman B will be at distance 1.5 meters in the opposite direction

The distance between the the two policemen is 1.5 + 1.5 = 3

Input:U = 5, V = 2, X = 4

Output:0.571429

**Approach:** It can be solved using **distance = speed * time**. Here, distance would be equal to the given range i.e. **distance = X** and speed would be the sum of the two speeds because they are moving in the opposite direction i.e. **speed = U + V**.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to return the time for which ` `// the two policemen can communicate ` `double` `getTime(` `int` `u, ` `int` `v, ` `int` `x) ` `{ ` ` ` `double` `speed = u + v; ` ` ` ` ` `// time = distance / speed ` ` ` `double` `time` `= x / speed; ` ` ` `return` `time` `; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `double` `u = 3, v = 3, x = 3; ` ` ` `cout << getTime(u, v, x); ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## Java

`// Java implementation of the approach ` `class` `GFG ` `{ ` ` ` ` ` `// Function to return the time for which ` ` ` `// the two policemen can communicate ` ` ` `static` `double` `getTime(` `int` `u, ` `int` `v, ` `int` `x) ` ` ` `{ ` ` ` `double` `speed = u + v; ` ` ` ` ` `// time = distance / speed ` ` ` `double` `time = x / speed; ` ` ` `return` `time; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` `int` `u = ` `3` `, v = ` `3` `, x = ` `3` `; ` ` ` `System.out.println(getTime(u, v, x)); ` ` ` `} ` `} ` ` ` `/* This code contributed by PrinciRaj1992 */` |

*chevron_right*

*filter_none*

## Python3

`# Python3 implementation of the approach ` ` ` `# Function to return the time ` `# for which the two policemen ` `# can communicate ` `def` `getTime(u, v, x): ` ` ` ` ` `speed ` `=` `u ` `+` `v ` ` ` ` ` `# time = distance / speed ` ` ` `time ` `=` `x ` `/` `speed ` ` ` `return` `time ` ` ` `# Driver code ` `if` `__name__ ` `=` `=` `"__main__"` `: ` ` ` ` ` `u, v, x ` `=` `3` `, ` `3` `, ` `3` ` ` `print` `(getTime(u, v, x)) ` ` ` `# This code is contributed ` `# by Rituraj Jain ` |

*chevron_right*

*filter_none*

## C#

`// C# implementation of the approach ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` ` ` `// Function to return the time for which ` ` ` `// the two policemen can communicate ` ` ` `static` `double` `getTime(` `int` `u, ` `int` `v, ` `int` `x) ` ` ` `{ ` ` ` `double` `speed = u + v; ` ` ` ` ` `// time = distance / speed ` ` ` `double` `time = x / speed; ` ` ` `return` `time; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `Main() ` ` ` `{ ` ` ` `int` `u = 3, v = 3, x = 3; ` ` ` `Console.WriteLine(getTime(u, v, x)); ` ` ` `} ` `} ` ` ` `// This code is contributed ` `// by Akanksha Rai ` |

*chevron_right*

*filter_none*

## PHP

**Output:**

0.5

## Recommended Posts:

- Number of ways to arrange K different objects taking N objects at a time
- Calculate speed, distance and time
- Find the distance covered to collect items at equal distances
- Distance of chord from center when distance between center and another equal length chord is given
- Convert characters of a string to opposite case
- Position of a person diametrically opposite on a circle
- Find two vertices of an isosceles triangle in which there is rectangle with opposite corners (0, 0) and (X, Y)
- Program to find simple moving average
- Check if it is possible to return to the starting position after moving in the given directions
- Maximum possible intersection by moving centers of line segments
- Time difference between expected time and given time
- Count possible moves in the given direction in a grid
- Exterior angle of a cyclic quadrilateral when the opposite interior angle is given
- Check if it is possible to draw a straight line with the given direction cosines
- Changing One Clock Time to Other Time in Minimum Number of Operations

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.